To solve \((\sqrt{2} + 1)^4\), we can use the binomial theorem, which states that:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

Here, \(a = \sqrt{2}\), \(b = 1\), and \(n = 4\). Applying the binomial theorem, we have:

\[
(\sqrt{2} + 1)^4 = \sum_{k=0}^{4} \binom{4}{k} (\sqrt{2})^{4-k} (1)^k
\]

This simplifies to:

\[
(\sqrt{2} + 1)^4 = \binom{4}{0} (\sqrt{2})^4 (1)^0 + \binom{4}{1} (\sqrt{2})^3 (1)^1 + \binom{4}{2} (\sqrt{2})^2 (1)^2 + \binom{4}{3} (\sqrt{2})^1 (1)^3 + \binom{4}{4} (\sqrt{2})^0 (1)^4
\]

Calculating each term separately:

1. \(\binom{4}{0} (\sqrt{2})^4 (1)^0 = 1 \cdot 4 \cdot 1 = 4\)
2. \(\binom{4}{1} (\sqrt{2})^3 (1)^1 = 4 \cdot 2\sqrt{2} \cdot 1 = 8\sqrt{2}\)
3. \(\binom{4}{2} (\sqrt{2})^2 (1)^2 = 6 \cdot 2 \cdot 1 = 12\)
4. \(\binom{4}{3} (\sqrt{2})^1 (1)^3 = 4 \cdot \sqrt{2} \cdot 1 = 4\sqrt{2}\)
5. \(\binom{4}{4} (\sqrt{2})^0 (1)^4 = 1 \cdot 1 \cdot 1 = 1\)

Summing these results:

\[
4 + 8\sqrt{2} + 12 + 4\sqrt{2} + 1 = 17 + 12\sqrt{2}
\]

So,

\[
(\sqrt{2} + 1)^4 = 17 + 12\sqrt{2}
\]

Would you like further details or have any questions?

Here are some related questions for further practice:

1. What is the expansion of \((\sqrt{2} - 1)^4\)?
2. How do you simplify \((\sqrt{3} + 1)^3\)?
3. What is the value of \((2 + \sqrt{3})^2\)?
4. How do you use the binomial theorem to expand \((x + y)^5\)?
5. What are the coefficients in the expansion of \((x - 1)^4\)?
6. How do you find the middle term in the expansion of \((a + b)^6\)?
7. What is the expansion of \((\sqrt{5} + 2)^3\)?
8. How do you simplify \((\sqrt{3} - 1)^3\)?

**Tip:** When working with the binomial theorem, it's often helpful to write out a few terms to identify patterns before calculating the entire expansion.

Learn how to solve the expression (sqrt(2) + 1)^4 using the binomial theorem. Step-by-step solution provided, including simplification of radical expressions and binomial coefficient calculations. Suitable for advanced high school students and anyone interested in algebraic expansions.

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